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In qualitative analysis of second group in lab, $H_2S$ gas is passed in acidic medium for ppt. When $Cu^{2+}$ and $Cd^{2+}$ react with KCN, then in which of the following conditions, ppt will not be formed (for Cu) due to relative stability :
A
$K_2[Cu(CN)_4]$ – more stable, $K_2[Cd(CN)_4]$ – less stable
B
$K_2[Cu(CN)_4]$ – less stable, $K_2[Cd(CN)_4]$ – more stable
C
$K_3[Cu(CN)_4]$ – more stable, $K_2[Cd(CN)_4]$ – less stable
D
$K_3[Cu(CN)_4]$ – less stable, $K_3[Cd(CN)_4]$ – more stable
Detailed Solution
In group II analysis, $Cu^{2+}$ and $Cd^{2+}$ are separated using KCN.
With excess KCN, copper forms the Cu(I) complex $K_3[Cu(CN)_4]$ (Cu is reduced from +2 to +1), and cadmium forms $K_2[Cd(CN)_4]$.
$[Cu(CN)_4]^{3-}$ is very stable, so the concentration of free $Cu^+$ is too low and no CuS (or $Cu_2S$) precipitates when $H_2S$ is passed.
$[Cd(CN)_4]^{2-}$ is less stable, so it gives enough $Cd^{2+}$ to form the yellow CdS precipitate.
So the condition is: $K_3[Cu(CN)_4]$ – more stable and $K_2[Cd(CN)_4]$ – less stable.
With excess KCN, copper forms the Cu(I) complex $K_3[Cu(CN)_4]$ (Cu is reduced from +2 to +1), and cadmium forms $K_2[Cd(CN)_4]$.
$[Cu(CN)_4]^{3-}$ is very stable, so the concentration of free $Cu^+$ is too low and no CuS (or $Cu_2S$) precipitates when $H_2S$ is passed.
$[Cd(CN)_4]^{2-}$ is less stable, so it gives enough $Cd^{2+}$ to form the yellow CdS precipitate.
So the condition is: $K_3[Cu(CN)_4]$ – more stable and $K_2[Cd(CN)_4]$ – less stable.